Solid Geometry

Help Questions

ACT Math › Solid Geometry

Questions 1 - 10

Find the volume of a cube with side length 10.

$10\sqrt{2}$

Explanation

To find volume, simply cube the side length. Thus,

Find the surface area of a cube with side length .

Explanation

To solve, simply multiply the face area by . Thus,

Find the surface area of a cube whose side length is .

Explanation

To find surface area of a cube, simply calculate the area of one side and multiply it by . Thus,

If the surface area of a cube equals 96, what is the length of one side of the cube?

Explanation

The surface area of a cube = 6a2 where a is the length of the side of each edge of the cube. Put another way, since all sides of a cube are equal, a is just the lenght of one side of a cube.

We have 96 = 6a2 → a2 = 16, so that's the area of one face of the cube.

Solving we get √16, so a = 4

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

Given the volume of a cube is , find the side length.

Explanation

To find side length, simply realize that volume is the side length cubed. Thus,

$s=\sqrt[3]{V}=\sqrt[3]{64}=\sqrt[3]{4^3}=4$

A cube has a surface area of $150\textup{m}^2$ . What is its volume?

$125\textup{m}^3$

$1000\textup{m}^3$

$225\textup{m}^3$

$300\textup{m}^3$

Explanation

First, find the side lengths of the cube.

Recall that the surface area of the cube is given by the following equation:

$\text{Surface Area}=6s^2$ , where is the length of a side.

Plugging in the surface area given by the equation, we can then find the side length of the cube.

Now, recall that the volume of a cube is given by the following equation:

$\text{Volume}=s^3$

$\text{Volume}=5^3=125$

Find the volume of a cube given side length is .

Explanation

To solve, simply cube the side length. Thus,

Find the surface of a isoceles pyramid whose height is , base side length is , and slant height is .

Explanation

To find surface area of a pyramid, simply find the surface area of each of the faces and add them together.

$\textup{Base}=s^2=6^2=36$

$\textup{Triangle face}=\frac{1}{2}bh=\frac{1}{2}*6*4=12$

Since there are 4 trangular faces, we have to multiply that surface area by 4. Thus,

Page 1 of 48

Return to subject

AI TutorYour personal study buddy

Question of the DayDaily practice to build your skills

FlashcardsStudy and memorize key concepts

WorksheetsBuild custom practice quizzes

AI SolverStep-by-step problem solutions

Learn by ConceptMaster concepts step by step

Diagnostic TestsAssess your knowledge level

Practice TestsTest your skills with exam questions

GamesLearn through free games